TITLE: Evaluate $\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right)$
QUESTION [7 upvotes]: I want to evaluate $$\lim_{n\to \infty}\left(\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+\cdots+\frac{1}{\sqrt{(2n-1)2n}}\right).$$
I have computed 
\begin{align*}
a_{n+1}-a_n & = \frac{1}{\sqrt{(2n+1)(2n+2)}}-\frac{1}{\sqrt{n(n+1)}}\\
            & = \frac{1}{\sqrt{n+1}}\left( \frac{1}{\sqrt{2(2n+1)}}-\frac{1}{\sqrt{n}} \right).
\end{align*}
Now I'm stuck.

REPLY [5 votes]: HINT
The hint provided by Chinnapparaj R is the key point, to evaluate the bounding sum we can use that
$$\frac1n+\frac1{n+1}+\ldots+\frac1{2n}=\sum_{k=1}^n \frac{1}{n+k}=\frac1n \sum_{k=1}^n \frac{1}{1+\frac kn}$$
which is a Riemann's sum.